Question : $ABC$ is a triangle. $AB = 5$ cm, $AC = \sqrt{41}$ cm and $BC = 8$ cm. $AD$ is perpendicular to $BC$. What is the area (in cm2) of $\triangle ABD$?
Option 1: 12
Option 2: 6
Option 3: 10
Option 4: 20
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 6
Solution :
Let the length of $BD=x$ cm, then $CD= (8-x)$ cm. In $\triangle ADB$ $AB^2=AD^2+BD^2$ ⇒ $5^2=AD^2+x^2$ ⇒ $AD^2=5^2-x^2$ ... (i) In $\triangle ADC$ $AC^2=AD^2+DC^2$ ⇒ $(\sqrt{41})^2=AD^2+(8-x)^2$ ⇒ $AD^2=41-(8-x)^2$ ... (ii) From (i) and (ii) $5^2-x^2=41-(8-x)^2$ ⇒ $25=-23+16x$ ⇒ $16x=48$ ⇒ $x=3$ From equation (i), $ AD^2=5^2-3^2$ ⇒ $AD=4$ cm Area of $\triangle ABD =\frac{1}{2}×AD×BD$ ⇒ Area of $\triangle ABD =\frac{1}{2}×4×3$ Area of $\triangle ABD = 6$ cm 2 Hence, the correct answer is 6.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : In a triangle ${ABC}, {AB}={AC}$ and the perimeter of $\triangle {ABC}$ is $8(2+\sqrt{2}) $ cm. If the length of ${BC}$ is $\sqrt{2}$ times the length of ${AB}$, then find the area of $\triangle {ABC}$.
Question : The centroid of a $\triangle ABC$ is G. The area of $\triangle ABC$ is 60 cm2. The area of $\triangle GBC$ is:
Question : In $\triangle$ABC, D and E are points on the sides BC and AB, respectively, such that $\angle$ACB = $\angle$ DEB. If AB = 12 cm, BE = 5 cm and BD : CD = 1 : 2, then BC is equal to:
Question : Suppose $\triangle ABC$ be a right-angled triangle where $\angle A=90°$ and $AD\perp BC$. If the area of $\triangle ABC =40$ cm$^{2}$ and $\triangle ACD =10$ cm$^{2}$ and $\overline{AC}=9$ cm, then the length of $BC$ is:
Question : In $\triangle$ABC, $\angle$BAC = 90º and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm, then the length of BC is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile